Thermodynamics

JEE Main level, numerical answer, medium

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Question 17 of 190

55 moles of unknown gas is heated at constant volume from 10 °C10 \, °C to 20 °C20 \, °C. The molar specific heat of this gas at constant pressure cp=8c_p = 8 cal/mol.°C and R=8.36R = 8.36 J/mol.°C. The change in the internal energy of the gas is _______ calorie.

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Answer: 300

Given n=5n = 5 moles, ΔT=20∘C−10∘C=10∘C\Delta T = 20^{\circ}\text{C} - 10^{\circ}\text{C} = 10^{\circ}\text{C}.

The molar specific heat at constant pressure is cp=8 cal/mol⋅∘Cc_p = 8 \text{ cal/mol}\cdot^{\circ}\text{C}.

The universal gas constant is R=8.36 J/mol⋅∘CR = 8.36 \text{ J/mol}\cdot^{\circ}\text{C}.

Converting RR into calories, we use 1 cal=4.18 J1 \text{ cal} = 4.18 \text{ J}:

R=8.364.18 cal/mol⋅∘C=2 cal/mol⋅∘CR = \dfrac{8.36}{4.18} \text{ cal/mol}\cdot^{\circ}\text{C} = 2 \text{ cal/mol}\cdot^{\circ}\text{C}

Using Mayer's relation, the molar specific heat at constant volume is:

cv=cp−R=8−2=6 cal/mol⋅∘Cc_v = c_p - R = 8 - 2 = 6 \text{ cal/mol}\cdot^{\circ}\text{C}

The change in internal energy of the gas is given by:

ΔU=ncvΔT\Delta U = n c_v \Delta T

ΔU=5×6×10=300 cal\Delta U = 5 \times 6 \times 10 = 300 \text{ cal}

Answer: 300300

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